Frattini subset

Suppose A is a set with a binary operation. Then we say a subset S of AgeneratesA if finite iterations of products (commonly called words over S) from the set Seventually produce every element of A. We write
this property as A=⟨S⟩.

Example 2.

Proof.

Take n≠0. Without loss of generality, n is positive. Then take
1<m some integer relatively prime to n. Thus by the
Euclidean algorithm we know there are integers a,b such that 1=a⁢m+b⁢n.
This shows 1∈⟨m,n⟩ so in fact {m,n} generates
ℤ.

However, ℤ is not generated by m, as m>1. Therefore n cannot be removed form the generating set{m,n} and so indeed the only non-generator of ℤ is 0.
∎

Example 3.

In the ring Z4, the element 2 is a non-generator.

Proof.

Check the possible generating sets directly.
∎

Example 4.

The set of positive integers under addition, N, has no non-generators.

Proof.

So we see not all sets with binary operations have non-generators. In the case that a binary operation has an identity then the identity always serves as a non-generator due to the convention that the empty word be defined as the identity. However, without further assumptions on the product, such as associativity, it is not always possible to treat the Frattini subset as a subobject in the category of the orignal object. For example, we have just shown that the Frattini subset of a semi-group need not be a semi-group.

Theorem 6.

Proof.

For a group G, given a non-generator a, and M any maximal subgroup
of G. If a is not in M then ⟨M,a⟩ is a larger subgroup
than M. Thus G=⟨M,a⟩. But a is a non-generator so
G=⟨M⟩=M. This contradicts the assumption that M is a maximal
subgroup and therefore a∈M. So the Frattini subset lies in every
maximal subgroup.

In contrast, if a is in all maximal subgroups of G, then given any
subset S of G for which G=⟨S,a⟩, then set M=⟨S⟩. If M=G, then a is a non-generator. If not, then M lies
in some maximal subgroup H of G. Since a lies in all maximal subgroup,
a lies in H, and thus H contains ⟨S,a⟩=G. As H is
maximal, this is impossible. Hence G=M and a is a non-generator.
∎